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# 3. Standing Waves

• • Wendell Potter and David Webb et al.
• Physics at UC Davis

So far we have restricted our discussion of waves to waves that travel. In all our examples until this part, one could follow the location of a maximum of the wave, for example, and observe it moving with the rest of the disturbance.

Another important class of waves exist called standing waves. For a standing wave, the position of the maximum and minima do not travel, but remain in place. There are many real-world standing waves; you may have noticed standing waves when you wiggled one end of a string, slinky, rope, etc while the other end was held fixed.

We will begin our discussion of standing waves by noting what occurs experimentally when standing waves form. Initially, it may be unclear why standing waves fit into our wave unit. After all, we emphatically emphasized that waves required propagation of a disturbance. Once we establish the idea of standing waves, we will use our model of interference to make sense of them.

# What is a Standing Wave?

In the most general sense, we have already defined a standing wave as a wave that does not travel. How do these waves come to exist? Imagine you have a string attached at both ends that is under tension, like a guitar string. If you try to vibrate it at a particular rate, you may or may not be successful depending on the frequency you choose. At most frequencies, the wave you start will travel to the one end intact, but upon reaching it the shape of the wave distorts and overall the string no longer appears to carry a wave. Nowhere will the string displace very far from equilibrium.

Only at certain frequencies will you see a sizeable displacement. If you begin vibrating at an extremely low frequency and gradually increase the frequency, the first wave-like response of the string will look like the image below. Each of the seven lines in the image is like a photograph of the string at a particular instant. These serve as displacement versus position graphs of the string, at seven different times. Notice that the displacement at both ends is zero. This makes a great deal of sense, because both ends are attached and thus cannot move. Any part of a standing wave that experiences no displacement over time is called a node; the above picture features two nodes. Also notice there is one spot in the middle that experiences the maximum amount of displacement at each time. Any spot that exhibits this behavior will be called an antinode

If we increase the frequency of our vibrations, we lose the wave shape for a while, but it returns at higher frequency values. The next three frequencies we find to exhibit wave-like behavior will give us the waves shown below. In each image, the arrows highlight a distance of a half wavelength. If we use $$L$$ to denote the length of the string, then for the first frequency, $$\lambda = 2L$$ because only half of a wavelength fits on the string. For the second lowest frequency, $$\lambda = L$$ since an entire wavelength fits on the string. Similar relationships can be established for $$n = 3$$ and $$n = 4$$. In general, for waves on a string that are attached at both ends,

$\lambda = \dfrac{2L}{n}$

Here, the various $$n$$ values specify which harmonic we are discussing. The lowest harmonic, with $$n = 1$$, is called the fundamental harmonic.

We have now developed a relationship between harmonic $$n$$ and wavelength $$\lambda$$. If we knew the wave speed on the string, we could determine the frequency we have produced using $$v_{wave} = \lambda f = f \frac{2L}{n}$$.

Because frequency does not change between media, whatever frequency is produced on the string is reproduced in the air and eventually makes it into our ear. It is the frequency that we hear as a particular note. To make sure the instrument plays the correct note, a musician must first tune the instrument so it produces the correct frequency. To do this the musician can change the tension that the string is under (which adjusts the wave speed, from $$v_{wave} = \sqrt{\mu / \tau}$$.

While playing the guitar a guitar player chooses different notes by putting fingers on the fretboard, which changes the effective length of the string (and lowers the frequency, from $$v_{wave}= f \frac{2L}{n}$$). When we name the note the instrument plays as a single frequency, such as 440 Hz, we're referring to the fundamental harmonic of that note. In general, musical instruments produce many harmonics when playing any particular note. The different combinations of harmonics give instruments different sounds, allowing people to differentiate between pianos, guitars, and violins even though all three instruments use vibrating strings.

We have now explored standing waves with both ends attached in some detail. A similar analysis could be done for standing waves with only one end attached and the other end free. In this case, the attached end will still behave like a node, but now the free end will behave like an antinode. For instance, only one quarter ($$\lambda /4$$) of a wave will fit along the length of the string for the fundamental frequency.

Exercise

Draw the first four harmonics for a wave with one end attached and one end free. Can you determine a general relationship between the string length and the wavelength?

# Applying the Interference Model

Now that we have some sense of what standing waves are, it is time to make sense of them. There are two independent ways of making sense of this phenomenon in terms of the interference that we've seen in Superposition Basics and Superposition of Harmonic Waves. We will explore both briefly. Both methods involve waves travelling down the medium in opposite directions and interfering along the way.

### Reflection at the Boundary

First, we send a continuous periodic wave down our medium, it hits the boundary at the end, and reflects. We now have two waves in the medium: the wave we are creating and the reflected wave. The two waves have the same frequency, determined by us, the source of the waves. Since both waves travel in the same medium, they have the same speed and thus the same wavelength as well. Because the waves travel in opposite directions, the position components of their equations have opposite signs. Mathematically, we have

$y_1 (x,t) = A \sin \left( \dfrac{2 \pi t}{T} - \dfrac{2 \pi x}{\lambda} + \phi_1 \right)$

$y_2 (x,t) = A \sin \left( \dfrac{2 \pi t}{T} + \dfrac{2 \pi x}{\lambda} + \phi_2 \right)$

Notice that we left subscripts on the phase constant term, $$\phi$$ . When the wave hits the end of the medium, two different things can happen to the phase constant. In one case, called a soft reflection, the phase constant remains unchanged and $$\phi_2 = \phi_1$$. In the other case, called a hard reflection, the phase constant of the reflected is completely out of phase with the phase constant of the incoming wave, so $$\phi_2 = \phi_1 + \pi$$.

The gif below helps to visualize what we just described. The blue wave is the wave we've created and the green wave is the wave reflecting back towards us. The red wave is the actual displacement of the rope; it's obtained by superposing the green and blue waves. The free end of the rope, the end we were initially moving to create the waves, is a location of a constructive interference. This location has constructive interference for all times. The rope itself alternates between a maximum, flat, and a minimum, but the interference is always constructive. This might seem counterintuitive at first, but thinking back to our introduction to wave interference, there can several places where the displacement of both waves is zero (because the total phase $$\Phi = 0$$ or (2 \pi\) etc.), so the total displacement of the sum is also zero. Nonetheless, this is a spot of constructive interference.

## Infinite Interference Wave

There is a second way to apply the interference model to understand standing waves. In this case, we do not imagine any reflections, so do not need to worry about whether a boundary change is ‘hard’ or ‘soft.’ Unfortunately, it is a bit more abstract.

Imagine there are infinite waves traveling in opposite directions. Although we will be thinking about a small section of medium, like a length of string, we imagine that the waves extend beyond the medium in question. Throughout all of space, these waves are interfering. Their interference is like a giant standing wave in all space that has nodes and antinodes in every direction, with no ends.

To use this idea, we think about the particular type of interference we have (like both ends fixed, or node-node interference). We then consider only the portion of the total interference pattern that meets these restrictions, apply this to our specific phenomenon, and ignore the rest.

Example 1

Determine if reflection at the free end of a rope is a hard reflection or a soft reflection

Solution

To visualize the phenomenon better, let’s first sketch the situation. We could choose any harmonic, and any behavior for the other end we want. The sketch chosen above shows the fundamental with one end attached and one end free, at five different times. At the free end of a rope, there is an antinode. At some specific times, the free end has a maximal displacement. In general, at any given instant in time, the free end has more displacement than any other part of the rope. We will keep this in mind.

As with any interference problem, there are three terms to consider that
might cause interference.

1. The time term. In this case, the waves in question have the same period. There will be no total phase difference contribution from the time term.
2. The spatial term. We are examining the wave just as it turns around. Neither wave has traveled further than the other wave. There will be no total phase difference contribution from the spatial term.
3. The phase constant term. We seek to determine what the phase constant might be. We know the choices for $$\Delta \phi$$ are 0 (soft reflection) or $$\pi$$ (hard reflection).

At this point, we know that the interference at the free end is entirely from the difference in phase constant, so $$\Delta \Phi = \Delta \phi$$. We either have $$\Delta \Phi= 0$$ or $$\Delta \Phi = \pi$$. In the first case, we would have constructive interference, and in the
second case destructive interference. Clearly we are not seeing destructive interference, but we are seeing constructive. Thus, the reflection at a free end must be a soft reflection with $$\Delta \phi = 0$$.